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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Show that each of the given three vectors is a unit vector: $ \large\frac{1}{7}$$ ( 2\hat i + 3\hat j + 6\hat k),\large\frac{1}{7}$$ ( 3\hat i - 6\hat j + 2\hat k), \large\frac{1}{7}$$ ( 6\hat i + 2\hat j - 3\hat k) $ Also, show that they are mutually perpendicular to each other.

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Toolbox:
  • The magnitude of a unit vectors is 1.
  • $ | x\hat i + y\hat j + z\hat k | = \sqrt{x^2+y^2+z^2}$
  • Two vectors are perpendicular if their dot product is zero
Step 1:
Let $\overrightarrow a=\large\frac{1}{7}$$(2\hat i+3\hat j+6\hat k)$
$\quad\overrightarrow b=\large\frac{1}{7}$$(3\hat i-6\hat j+2\hat k)$
$\quad\overrightarrow c=\large\frac{1}{7}$$(6\hat i+2\hat j-3\hat k)$
Let us find the magnitude of $\overrightarrow a,\overrightarrow b$ and $\overrightarrow c$.
$|\overrightarrow a|=\sqrt{\big(\large\frac{2}{7}\big)^2+\big(\large\frac{3}{7}\big)^2+\big(\large\frac{6}{7}\big)^2}$
$\quad\;\;=\large\frac{1}{7}$$\sqrt{4+9+36}$
$\quad\;\;=\large\frac{7}{7}$$=1$
Similarly $|\overrightarrow b|=\sqrt{\big(\large\frac{3}{7}\big)^2+\big(\large\frac{-6}{7}\big)^2+\big(\large\frac{2}{7}\big)^2}$
$\qquad\qquad\quad=\large\frac{1}{7}$$\sqrt{9+36+4}$
$\qquad\qquad\quad=\large\frac{7}{7}$=1.
Similarly $|\overrightarrow c|=\sqrt{\big(\large\frac{6}{7}\big)^2+\big(\large\frac{2}{7}\big)^2+\big(\large\frac{-3}{7}\big)^2}$
$\qquad\qquad\quad=\large\frac{1}{7}$$\sqrt{36+4+9}$
$\qquad\qquad\quad=\large\frac{7}{7}$=1.
Therefore $\overrightarrow a,\overrightarrow b$ and $\overrightarrow c$ are unit vectors.
Step 2:
Next let us prove that they are mutually perpendicular to each other.
$\overrightarrow a.\overrightarrow b=\large\frac{1}{7}$$(2\hat i+3\hat j+6\hat k).\large\frac{1}{7}$$(3\hat i-6\hat j+2\hat k)$
$\qquad=\large\frac{1}{49}$$\big[(2\times3)+(3\times-6)+(6\times2)\big]$
$\qquad=\large\frac{1}{49}$$(6-18+12)=0$
$\overrightarrow b.\overrightarrow c=\large\frac{1}{7}$$(3\hat i-6\hat j+2\hat k).\large\frac{1}{7}$$(6\hat i+2\hat j-3\hat k)$
$\qquad=\large\frac{1}{49}$$\big[(3\times6)+(-6\times2)+(2\times-3)\big]$
$\qquad=\large\frac{1}{49}$$(18-12-6)=0$
$\overrightarrow a.\overrightarrow c=\large\frac{1}{7}$$(2\hat i+3\hat j+6\hat k).\large\frac{1}{7}$$(6\hat i+2\hat j-3\hat k)$
$\qquad=\large\frac{1}{49}$$\big[(2\times6)+(3\times2)+(6\times-3)\big]$
$\qquad=\large\frac{1}{49}$$(12+6-18)=0$
$\Rightarrow$ All the three vectors are mutually perpendicular.
answered May 20, 2013 by sreemathi.v
 

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